I got this problem:

Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$.
(Hint: the solution involves limits at infinity)

I tried to prove it (by contradiction) but I failed. Thanks on any help.


3 Answers 3


DISCLAIMER: This answer is incorrect as it stands, namely in the $k>0$ section is claims that: $$ f(0)=f(x-x)\geq f(x)-x f(f(x)) $$ implies $f(x)<f(0)$, which is false.

This disclaimer will stay here until the problem is fixed.

From the condition $f(0)>0$ it follows that with $x=0$ we have $$ f(0+y)-y f(f(0))\geq f(0)>0 $$ so it follows that $f(y)>ky$ where $k=f(f(0))$ is some non-zero constant. It has to be non-zero since otherwise $0=k=f(f(x))>k f(x)=0$. Now we have to divide into cases.

If $k>0$

If $k>0$ it follows that $x f(f(x))>kx f(x)>k^2 x^2\geq 0$. So in that case we have $$ \underbrace{f(x-x)}_{f(0)}\geq f(x)-\underbrace{x f(f(x))}_{\text{strictly greater than }0}\\ \iff\\ f(x)<f(0) $$ But the above statement is absurd by all means. For one thing $f(0)<f(0)$ and moreover $f(0)>f(x)>kx\rightarrow\infty$ which is also absurd.

If $k<0$

If we apply the beginning of what Denis showed, namely that $f(x)\rightarrow\infty$ for $x\rightarrow\infty$, then we get in addition to that from my analysis that for $k<0$ we have $f(-x)\geq -kx\rightarrow\infty$ for $x\rightarrow\infty$. So we see that $$ f(0)=f(-x+x)\geq f(-x)+x f(f(-x))\rightarrow\infty $$ for $x\rightarrow\infty$ too. A contradiction since we have just shown that the constant $f(0)$ is greater than or equal to an expression that tends to infinity. So $k<0$ is impossible too.

  • $\begingroup$ If you put x = 0, then the LHS in the first inequality you gave should be f(0) - yf(f(0)) $\endgroup$ Sep 24, 2014 at 8:43
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    $\begingroup$ @TenaliRaman: No! The LHS is still $f(0+y)-y f(f(0))$ even if $y=0$, and the statement $f(y)>ky$ holds for ANY $y$ including $y=0$. $\endgroup$
    – String
    Sep 24, 2014 at 8:49
  • $\begingroup$ @TenaliRaman: But I spotted another serious problem in my analysis which I just pointed out in my answer above ... $\endgroup$
    – String
    Sep 24, 2014 at 8:53
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    $\begingroup$ @String: Maybe we can suppose that $\forall x\in\Bbb{R}, f(f(x))\leq 0$ and from that reach to a contradiction and get that $\exists z\in\Bbb{R}$ such that $f(f(z))>0$ which will be something like the $k$ that you mentioned. $\endgroup$
    – MathNerd
    Sep 24, 2014 at 9:04
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    $\begingroup$ @Saita: I finished my argument! $\endgroup$
    – String
    Sep 24, 2014 at 10:32

By what Denis said we get that $lim_{x\to\infty}f(x)=\infty$ And so $lim_{x\to\infty}f(f(x))=\infty$.

Now we will reach a contradiction by using the definition of limits at infinity:

Because $lim_{x\to\infty}f(x)=\infty$ and $lim_{x\to\infty}f(f(x))=\infty$. We get that:
$\forall 0<M_1,\exists 0<N_1$ such that $\forall N_1<x, M_1<f(x)$ and that
$\forall 0<M_2,\exists 0<N_2$ such that $\forall N_2<x, M_2<f(f(x))$

Now take $M_1=M_2=1$ and we will get that $\exists 0<N_1$ such that $\forall N_1<x, 1<f(x)$ and that $\exists 0<N_2$ such that $\forall N_2<x, 1<f(f(x))$

Take $x=max (N_1+1,N_2+1)$, then $N_1,N _2<x$ and so we get that $1<f(x)$ and that $1<f(f(x))$

Now take $y=max(\frac{x+1}{f( f( x))-1},N_2-x)$ which implies that $N_2< x+y+1\leq y f(f(x ))$ and so $1<f(f(x+y+1))$, Now $f (x+y)\geq f(x )+yf(f(x))\geq x+y+1$ And hence

$f(f(x+y))\geq f(x+y+1)+(f(x+y)-(x+y+1))f(f(x+y+1)) \geq f(x +y+1)\geq f(x+y)+f(f(x+y))\geq f(x)+yf(f(x))+f(f(x+y))>f(f(x+y))$

And so we get that $0>0$ which is a contradiction.


Let $a=f(0)>0$, $b=f(a)$ and $c=f(b)$.

We know that $f(0+b)\geq a+b^2$, so we get $c>0$.

We also have for all $x$, $f(x+a)\geq f(a)+cx$, so in particular $\lim_{x\to\infty} f(x)=\infty$ and more precisely $f(x)=\Omega(x)$ when $x\to\infty$.

This means that for $x$ big enough, $f(x)>0$, and therefore, $f(x+1)\geq f(x)+f(f(x))>f(f(x))$. This gets us $f(f(x-1))<f(x)$.

Moreover, for $x$ big enough, we have $f(x-1)>x$ (since $f(x+1)=\Omega(x^2)$).

Let us choose $x_0$ big enough, and define $x_{n+1}=f(x_n-1)$.

The sequence $(x_n)$ is strictly increasing, and for all $n$, we have $f(x_{n+1})<f(x_n)$. We reached a a contradiction with $\lim_{x\to\infty} f(x)=\infty$.

  • $\begingroup$ So to sum up you have shown $f(x+a)\geq cx+b$ where $c>0$. Very nice! How do you conclude from that? $\endgroup$
    – String
    Sep 24, 2014 at 9:29
  • $\begingroup$ I found the end :) $\endgroup$
    – Denis
    Sep 24, 2014 at 10:11
  • $\begingroup$ It took me a while to absorb, but it makes sense now. Perhaps you should state the inequalities as $f(x)>f(f(x-1))$ and $f(x-1)>x$ respectively to make the reading more direct, but that is just an opinion. The first part of your answer showing that $f(x)\rightarrow\infty$ provided the last piece of the puzzle for my version of it BTW :) $\endgroup$
    – String
    Sep 24, 2014 at 10:40
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    $\begingroup$ I smoothed it accordingly. $\endgroup$
    – Denis
    Sep 24, 2014 at 10:47
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    $\begingroup$ I'm not sure I see where you get that $f(x+1) = \Omega(x^2)$. And also, the sequence $(x_n)$ may be strictly increasing without actually approaching infinity, so the contradiction seems premature. $\endgroup$ Jul 10, 2022 at 16:50

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